Notes
a square, triangle, and circle in a square solution

A Square, Triangle, and Circle in a Square

Solution by Pythagoras' Theorem

Let $a$, $b$, $c$ be the lengths of the sides of triangle $A B C$, so that $A B$ has length $a$, $B C$ of length $b$, and $A C$ of length $c$. Since the central square has area $100$, $c = 10$. The area of the triangle is given by $\frac{1}{2} a b$ so $a b = 48$. Applying Pythagoras' theorem to this triangle shows that $a^2 + b^2 = 100$. Putting these together, $(a + b)^2 = 100 + 2 \times 48 = 196$, so $a + b = 14$.

Triangle $C E G$ is congruent to triangle $A B C$. Inside that triangle, $C D$ and $C H$ have the same length, as to $G H$ and $G F$. So the combined lengths of $F E$ and $E D$ give the difference between the combination of $C E$ and $E G$ and the length of $C G$. So $F E$ has length $\frac{14 - 10}{2} = 2$ and so the area of the circle is $4 \pi$.